Constant $g$ acceleration from astronaut's frame of reference When a spaceship is experiencing a constant acceleration of $10m/s^2$, the astronauts will be moving at nearly the speed of light after about a year in the earth's reference frame. This means the spaceship's energy will start to diverge as a function of the speed $v$ so there will be a huge amount of energy necessary to increase the speed of the ship any further. This way, the speed of light can never be crossed.
All of this is clear to me, but all of this is also formulated in earth's reference frame. But from the astronaut's reference frame: the spaceship is simply accelerating at $10m/s^2$ and so the force on the spaceship is constant. Then why would we need huge amounts of energy to accelerate the spaceship?
For example, I read somewhere that the amount of energy that would be needed to accelerate a large spaceship to half the speed of light is more than 2000 times the current world annual energy consumption. How does this make sense in the astronaut's (non-inertial) frame?
 A: 
But from the astronaut's reference frame: the spaceship is simply accelerating at 10m/s2 and so the force on the spaceship is constant. Then why would we need huge amounts of energy to accelerate the spaceship?

I think the astronaut interprets the rocket's need for a huge amount of energy in another way. He observes that as he recedes from the earth, more amount of energy is needed than before to make the distance slightly greater. When the observer on earth measures the rocket's speed very close to light speed, the astronaut observes that the earth no longer recedes from him, and it just becomes red-shifted till it completely disappears. (The distance asymptotes to $c^2/a$.) In this case, the astronaut admits that for making an infinitesimally small displacement from the earth, the rocket's engine must be "on" for a very long time that may justify the need for huge energy.
See "Below the rocket, something strange is happening..." at this link. 
A: In the rocket, it seems that the amount of energy per unit time stays constant. When the astronauts look outside they see al the other objects in the universe move faster and faster (it seems these objects are in free fall in a homogeneous gravitational field). This means that the astronauts see the time on these objects move slower and slower. 
So say, for example, that when one second has passed in the ship, half a second has passed on all the other objects. The astronauts conclude that for all these objects the spaceship uses in this case twice as much energy per unit time. 
The spaceship speeds up. Then there comes a moment that the astronauts see that the time on all the other objects goes at a pace that's 1/3 of the time in the ship. So the astronauts (who still use the same amount of energy per unit time) that for all these objects the spaceship uses three times as much energy because in one unit of time on these objects three units of time are used in the spaceship. 
The spaceship approaches lightspeed. The astronauts (for whom the amount of energy used per unit time still is the same) see that the pace of time on all the other objects approaches zero. This means that the astronauts conclude that for all these objects the amount of energy per unit time used in the ship approaches infinity.
Of course, it's only the spaceship that accelerates and who's (relativistic) kinetic energy is increased. Here the twin paradox springs to mind. It's the spaceship that's first accelerating. After it has stopped accelerating, the universe and the spaceship are in relative motion to each other. If we let the spaceship return to Earth then the astronauts will be much younger as the people on Earth. This is an asymmetric situation. If the whole universe accelerates towards the spaceship then the people on Earth (on arriving near the spaceship) will have the same age as the astronauts in the ship. This is a symmetric situation. But that aside.
A: When pondering relativistic effects it can help to consider them from a reciprocal perspective. 
Suppose I am on the ship, and my speed has reached 0.9c, say, relative to Earth. At that point I cut my engines and coast past you on Earth. In my frame I am at rest and it is you who are racing by at 0.9c. You get into your car and accelerate down a road away from me to reach a speed of 60mph. To you, the acceleration is entirely normal, and uses the same amount of fuel etc--the fact that you were moving at 0.9c relative to me makes no difference in your frame. However, your speed relative to me is not 0.9c plus 60mph, as we have to use relativistic velocity addition, so it is somewhat less. The result is that your acceleration relative to me is less. 
If we repeated this scenario when I was coasting past at 0.999999999999999999c, the result of your acceleration to 60mph on Earth would be a near negligible increase in our overall relative speed.
So the answer is that if the effect of a constant acceleration in Frame A is a given increase in velocity v, say, then owing to the relativistic addiction of velocity, the magnitude of v approaches zero in Frame B as the relative speed of the two frames approaches c. The constant acceleration uses energy at a constant rate in Frame A to diminishing overall effect in Frame B.
A: From the rocket's frame of reference, the rocket is at rest and the Earth is travelling faster and faster, approaching c. In both frames of reference, the relative velocity approaches c so the energy needed diverges.
I'm not sure whether this answers your question.
A: If you know how much fuel (mass)  you need to travel from the earth to the next object , you can answer the question how much energy you need , remember that energy is  equivalent to mass.
I found the answer in this dokument   
http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html
$$\frac{m_F}{m_L}=\exp\left({\frac{a\,T}{c}}\right)-1\tag 1$$
where:
$m_F$ is the fuel mass  
$m_L$ is the payload mass  
$a$ is the constant acceleration 
$T$ is the rocket time 
$c$ is the light velocity
Example:
for:
$a=1 g$
$g\approx 1.03 [ly/yr^2]$
$c=1 [ly/yr]$
if you want to go to a distance from the earth 
$d=4.3\,\, [ly]$ you need to travel  $T=3.6$ years
$$d=\frac{{c}^{2}}{a} \left( \cosh \left( {\frac {aT}{c}} \right) -1 \right) 
$$
thus equation (1)
for every kilogram payload ($m_L=1$) you need $m_F=10\,[kg]$ fuel . from here you can calculate the energy  $E_F=m_F\,c^2$
